I want to find the volume of an $n$-dimensional simplex, i.e. determine \begin{align*} \sigma _{n} = \int_{A_{n}}^{} \mathrm{~d}\mu , \quad A_{n}= \{ ( x_{1}, \ldots, x_{n})\in \mathbb{R}^{n} \mid \forall i\colon x_{i} \geqslant 0, \quad x_{1} + \ldots+ x_{n} \leqslant 1\} .\end{align*} I came up with the following informal computation: We consider \begin{align*} A' = [0, 1], \quad A_{x_{1}} = \{ ( x_{2}, \ldots, x_{n})\mid x_{2} + \ldots+x_{n} \leqslant 1- x_{1}\} \end{align*} and conclude \begin{align*} \sigma _{n} &= \int_{0}^{1} \left(\int_{A_{x_{1}}}^{} \mathrm{~d}\mu ( x_{2}, \ldots, x_{n}) \right)\mathrm{~d}x _{1} \\ &\stackrel{(*)}{=} \int_{0}^{1} \sigma _{n-1} (1 - x_{1})^{n-1}\mathrm{~d}x _{1} \\ &= \sigma _{n-1}\left[-\frac{1}{n}( 1 - x_{1})^n\right]_{0}^{1} = \frac{\sigma _{n-1}}{n} .\end{align*} Solving this recursion we find \begin{align*} \sigma _{1} = \int_{0}^{1} \mathrm{~d}x _{1} = 1 \implies \sigma _{n} = \frac{\sigma _{1}}{n!} = \frac{1}{n!} .\end{align*}
How I can formalise the $(*)$ step?
The interval can be iteratively splitted up, i.e. first you have \begin{align*} &A' = \{ ( x_{1}, \ldots, x_{n-1})\mid x_{1} + \cdots + x_{n-1}\leqslant 1\} \\ &A_{(x _{1}, \ldots, x_{n-1})} = \{ x_{n}\mid x_{n} \leqslant 1 - x_{1} - \cdots - x_{n-1}\} \end{align*} yielding \begin{align*} \int_{A_{n}}^{} \mathrm{~d}\mu &= \int_{A'}^{}\left( \int_{A( x_{1}, \ldots, x_{n-1})}^{} \mathrm{~d}x _{n} \right)\mathrm{~d}\mu ( x_{n-1}, \ldots, x_{1}) \\ &= \int_{A'}^{} \int_{0}^{1 - x_{1} - \cdots - x_{n-1}} \mathrm{~d}x _{n} \mathrm{~d}\mu ( x_{n-1}, \ldots, x_{1}) .\end{align*} Continuing this procedure with $A'$ one ends up with \begin{align*} \sigma _{n} = \int_{0}^{1} \int_{0}^{1 - x_{1}} \cdots \int_{0}^{1 - x_{1} - \cdots - x_{n - 1}} \mathrm{~d}x _{n} \cdots \mathrm{~d}x _{2}\mathrm{~d}x _{1} .\end{align*} To compute this, we write \begin{align*} f _{n}( r) = \int_{0}^{r} \int_{0}^{r - x_{1}} \cdots \int_{0}^{r - x_{1} - \cdots - x_{n - 1}} \mathrm{~d}x _{n}\cdots \mathrm{~d}x _{2} \mathrm{~d}x _{1} .\end{align*} This allows us to establish a recursion, namely \begin{align*} f _{n+1}( r) &= \int_{0}^{r} \cdots \int_{0}^{r - x_{1} - \cdots - x_{n}} \mathrm{~d}x _{n+1}\cdots \mathrm{~d}x _{1} = \int_{0}^{r} f _{n}( r - x_{1})\mathrm{~d}x _{1} \\ &= -\int_{r}^{0} f _{n}( x_{1}) \mathrm{~d}x _{1} = \int_{0}^{r} f _{n}( x_{1}) \mathrm{~d}x _{1} \\[5pt] &\implies f _{n+1}'( r) = f _{n}( r) .\end{align*} With the base case $ f _{1}( r) = r$ this yields \begin{align*} f _{n}( r) = \frac{r ^{n}}{n!} \implies f _{n}(1) = \frac{1}{n!} .\end{align*}